# Question #4c9b0

##### 1 Answer

#### Explanation:

In order to find a solution's **molality**, you must know two things

thenumber of moles of solutehow manykilograms of solventyou have in the solution

You're dealing with a **percent concentration by mass** to determine how many *grams* of solute and how many *grams* of solvent you get in a **random sample** of this solution.

To make the calculations easier, pick a **for every** **of solution**.

Use sodium chloride's **molar mass** to determine how many moles of solute you have in this sample

#2 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.03422 moles NaCl"#

Now, this solution will also contain

#m_"solution" = m_"solute" + m_"solvent"#

#m_"solvent" = "100 g" - "2 g" = "98 g water"#

This means that its molality will be - **do not** forget to convert the mass of water from *grams* to *kilograms*

#color(blue)(|bar(ul(color(white)(a/a)b = n_"solute"/m_"solvent"color(white)(a/a)|)))#

#b = "0.03422 moles"/(98 * 10^(-3)"kg") = "0.3492 mol kg"^(-1)#

I'll leave the answer rounded to two **sig figs**, despite the fact that you only have one sig fig for the percent concentration of the solution

#b = color(green)(|bar(ul(color(white)(a/a)"0.35 mol kg"^(-1)color(white)(a/a)|)))#

**SIDE NOTE** *The molality of the solution must be the same regardless of what sample you pick as a starting point.*

*I highly recommend redoing the calculations for different sample of this solution - the molality will always be equal to*